so the question I was trying to answer was, find the volume of the solid bounded above by the ellipsoid of revolution
$$b^2 x^2 + b^2 y^2 + a^2 z^2 = a^2 b^2,$$
below by the x-y plane
and on the sides by the cylinder $$x^2 + y^2 - ay = 0$$
I converted this into polar coordinates and got two possible double integrals to get my answer. The second one got me the correct answer, the first one got me the wrong answer. Could someone please help me figure out why the first integral calculates the wrong answer? It's been driving me mad for a day now!
Here is the one that gave me an incorrect answer (https://i.stack.imgur.com/EZ8GI.jpg)
and Here is the one that gave me the correct answer. (https://i.stack.imgur.com/MjQMr.jpg)
In the first one I tried to exploit the symmetry, the volume is the volume above the first quadrant in the xy plane multiplied by two.
Help would be very much appreciated. I guess I'll need to rack up some reputation points and put a bounty on this.. Not getting anywhere by myself
Possible problem: doing the inner integral: $$\int_0^{a\sin\theta}br\sqrt{1 - r^2/a^2}\,dr = -\frac{b}{3a}(a^2 - r^2)^{3/2}\Big\vert_{r = 0}^{r = a\sin\theta} = \frac{a^2b}3(1 - |\cos^3\theta|)$$ and the $|\dot{}|$ is essential.
EDIT: The symmetry is true. For $\theta\in[0,\pi/2]$ we have $|\cos\theta| = \cos\theta$ and: $$ \int_0^\pi\frac{a^2b}3(1 - |\cos^3\theta|)\,d\theta = 2\int_0^{\pi/2}\frac{a^2b}3(1 - |\cos^3\theta|)\,d\theta = 2\int_0^{\pi/2}\frac{a^2b}3(1 - \cos^3\theta)\,d\theta\ne \int_0^{\pi}\frac{a^2b}3(1 - \cos^3\theta)\,d\theta. $$